POJ 1141

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#define M 105
using namespace std;
const int inf = 0x13131313;
char s[M];
int dp[M][M] = {0};
int record[M][M];
int Min(int a,int b){return a>b?b:a;}
void print(int i,int j)
{
	if(i > j) return ;
	if(i == j) 
	{
		if(s[i] == '(' || s[i] == ')') printf("()");
		if(s[i] == '[' || s[i] == ']') printf("[]");
		return ;
	}
	if(record[i][j] == -1)
	{
		printf("%c",s[i]);
		print(i+1,j-1);
		printf("%c",s[j]);
	}
	else
	{
		print(i,record[i][j]);
		print(record[i][j]+1,j);
	}
}
int main()
{
	scanf("%s",s);
	int index;
	int n = strlen(s);
	for (int i = 0; i < n; ++i)
		dp[i][i] = 1;
	memset(record,inf,sizeof(record));
	for (int i = 1; i <= n-1; ++i) //间隔为1
	{
		for (int j = 0; j < n-i; ++j)
		{
			index = j + i;
			dp[j][index] = inf;
			/*
			其实，才开始这个想法就是看的矩阵连乘的dp规划。
			这题，我一直觉得有个弯子，比较难转过来，
			看了黑书，才慢慢明白了，
			dp[j][index] = Min(dp[j][index],dp[j+1][index-1]);
			比如dp[1][2] = Min(dp[1][2],dp[2][1]);
			发现才开始没有在dp[2][1]做了计算。不过这种情况，dp[1][2]=0;
			所以说，dp[i+1][i]=0;所以才开始的时候对dp数组赋值只对dp[i][i]赋值。
			*/
			if((s[j] == '(' && s[index] == ')') || (s[j] == '[' && s[index] == ']'))  
			{
				record[j][index] = -1;
				dp[j][index] = Min(dp[j][index],dp[j+1][index-1]);
			}	
			for (int k = j; k < index; ++k)
			{ 
				if(dp[j][k] + dp[k+1][index] < dp[j][index])
				{
					dp[j][index] = dp[j][k] + dp[k+1][index];
					record[j][index] = k;
				}
			}
		}
	}
	print(0,n-1);
	printf("\n");
	return 0;